Theory:

. In any triangle ABC the algebraic
sum of the distances from the circumcenter O
to the sides, is R+r, the sum of circumradius
and the inradius.

Solutions from the Previous Issue:

1.What is the maximum number for the possible points of intersection
of a circle and a triangle?

Solution:

There are 6 possible points of intersection of a
circle and a triangle.

2.Solve for x:

Solution:

The procedure is to multiply both sides of the
equation by the LCD, which in this case is (x  2)(x + 1). It is essential to
multiply every term on both sides of the equation. In so doing, all denominator
cancel.

x + 1 2(x  2) = 7

x + 1  2x + 4 = 7

-x = 2

x = -2

3.Solve for x:

Solution:

The LCD is (x  3)((x  2).

Multiplying both sides by this
yields

3(x - 2)  2(x  3) = 3

3x  6  2x + 6 = 3

x = 3

4.Solve for x:

Solution:

We multiply both sides by the LCD, (x  1)(x +
3), to obtain

x + 3 + 3(x  1) = (x  1)(x + 3)

x + 3 + 3x  3 = x2 + 2x 3

0 = x2 - 2x  3

(x 3)(x + 1) =0

x = 3; x = -1

Since the LCD is nonzero for each
of these, the solution set is {-1,3}